MARK SCHEME for the May/June 2012 question paper for the ...maxpapers.net/CIE/A Level/Mathematics – Further (9231)/9231_s12... · UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Level

MARK SCHEME for the May/June 2012 question paper

for the guidance of teachers

9231 FURTHER MATHEMATICS

9231/13 Paper 1, maximum raw mark 100

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper

GCE A LEVEL – May/June 2012 9231 13

© University of Cambridge International Examinations 2012

Mark Scheme Notes

Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.

Accuracy marks cannot be given unless the associated method mark is earned (or implied).

B Mark for a correct result or statement independent of method marks.

• When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given.

• The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working.

• Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.

• Wrong or missing units in an answer should not lead to the loss of a mark unless the

scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,

or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.




The following abbreviations may be used in a mark scheme or used on the scripts:

AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that

the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely

clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error

is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently

accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a

case where some standard marking practice is to be varied in the light of a particular circumstance)

Penalties

MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or

part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting.

PA –1 This is deducted from A or B marks in the case of premature approximation. The

PA –1 penalty is usually discussed at the meeting.




Qu

No

Commentary Solution Marks Part

Mark

Total

1 Finds partial fractions. Use method of differences. Obtains results.

+−=

+ 2

11

2

1

)2(

1

rrrr

=

+∑=

n

rrr1 )2(

1

−+

−++

+−

−+

+−

3

11

4

1

2

1...

1

1

1

1

2

11

2

1

nnnn

+−

+−=

2

1

1

1

2

3

2

1

nn

(acf) 4

3=⇒

∞S

M1A1

M1

A1A1√ 5 [5]

2 (States proposition.)

Proves base case.

States Inductive hypothesis.

Proves inductive step.

States conclusion.

)24

34 :P( −

=

n

nnu

Let n = 1 4 × 4

3 – 2 = 3 –2 = 1⇒P1 true.

Assume Pk is true for some k.

4

26

4

3.

4

3.4

4

224

343

1

+−

=

−

−

=+

k

k

ku

1

1

PP 24

3.4

+

+

⇒∴−

=

kk

k

∴ By PMI Pn is true ∀ positive integers.

B1

B1

M1

A1 A1 5 [5]

3 0

d

d1)(3

d

d 2 =

++++

x

yyx

x

yxy

0330 =′++′+ yy

4

3−=′⇒ y (AG)

0d

d)(3

d

d1)(6

d

d

d

d

d

d2

2

2

2

2

2

=++

+++++

x

yyx

x

yyx

x

yx

x

y

x

y

0316

16

4

3

4

3=′′+×+′′+−− yy

32

9=′′⇒ y

N.B. Mark similarly if expression expanded before differentiating.

B1B1

B1

B1 B1B1

M1

A1

3

5

[8]




Qu

No


Mark

Total

4 Integrates by parts. Obtains reduction formula. Finds I0 ( or I1) and uses reduction formula. (M1A1 for I1

if found immediately.) Obtains I2 . Obtains I3 .

∫=e

1

2d ln xx)(xI n

n

x

x

x

xn

x

xn

e

n d3

1)(ln

3)(ln

3e

1

1

1

3

∫ ⋅⋅−

= −

1

3

33

e

−

−=nI

n (AG)

∫−

=

==

e

1

3e

1

3

2

03

1e

3d

xxxI

9

1e2

3

1e

3

1

3

e333

1

+=

−−=⇒ I

27

2e5

9

1e2

3

2

3

e333

2

−=

+−=⇒ I

27

2e4

27

2e5

3

e333

3

+=

−−=⇒ I

M1A1 A1

A1

B1

M1

A1

A1

4

4 [8]

5 Proves initial result. Finds eigenvectors corresponding to given eigenvalues. Finds corresponding eigenvalue. Recognises the result proved initially. Gives eigenvalues

and matches eigenvectors.

(A + kI)e = Ae + kIe = λe + ke = (λ + k)e

∴ (A + kI) has eigenvalue (λ + k) with corresponding eigenvector e. Eigenvalues are –3 and 4 (given). Corresponding

eigenvectors are

−

1

1

1

,

0

1

1

Third eigenvalue is 6. C = B – 3I (Stated or implied.) Eigenvalues: –6, 1, 3 .

Corresponding eigenvectors are:

−

1

1

1

,

0

1

1

,

−

−

2

1

1

. (OE)

(For ‘non hence’ method, using characteristic equation, award B1 rather than M1A1 for eigenvalues, followed by B1 for eigenvectors.)

M1A1

M1A1 A1

B1

M1

A1

A1√

2

3

1

3

[2]

[3]

[1]

[3]




Qu

No


Mark

Total

6 States vertical asymptote.

Finds oblique asymptote. Differentiates and equates to zero. Finds x coordinates. States coordinates of turning points. Deduct at most 1 mark for poor forms at infinity.

Vertical asymptote is x = 2.

2

42

−

++=

xxy

Oblique asymptote is y = x + 2.

4)2(0)2(

41 2

2=−⇒=

−−=′ x

xy

x = 0, 4. Turning points are (0,0) and (4,8) Axes and both asymptotes correct. Upper branch correct. Lower branch correct.

B1

M1 A1

M1

A1

A1

B1 B1 B1

3

3

3 [9]

7 Complete strategy and getting halfway.

Fully correct.

Grouping.

Correct LHS and RHS.

Sets up substitution. Uses result obtained above.

Obtains result.

+−

++=

−

+

4

4

2

2

24

12

12

11

z

z

z

z

z

z

z

z

642

246 12142

zzz

zzz ++−−−+=

411

21

2

2

4

4

6

6 −

+−

++

+=

z

z

z

z

z

z

42cos24cos46cos2sin4cos16

24−−+=−⋅ θθθθθ

θθθθθ 6cos24cos42cos24sincos6424

−−+=

θθ

θ sin2d

d cos2 −==

x

x

02 3

1 =⇒==⇒= θπ

θ xx

∫ ∫=−

0

3

3

0

2424dsincos64dsin4.cos16

π

π

θθθθθθ

(LR – allow use of 2π, if seen.).

∫ −−+=3

0

d)6cos24cos42cos24(

π

θθθθ

3

0

6sin3

14sin2sin4

π

θθθθ

−−+=

33

4

2

3

2

3

3

4+=

++=

ππ

(AG)

M1A1

A1

M1

A1(L)

A1(R)

M1

M1

A1

A1

6

4 [10]




Qu

No


Mark

Total

8 (i)

(ii)

Deduces initial result. Substitutes into cubic equation. Deduces new cubic equation.

Deduces initial result.

Substitutes into cubic equation. Deduces new cubic equation.

12 =++=+⇒++−= γβααγβα uu

−=⇒

2

1 u

α

0102

13

2

1

2

123

=−

−−

−−

−⇒

uuu

⇒… 09313

23=+−−⇒ uuu

10=αβγ

v

v

v 1010

111=⇒==⇒=⇒ α

αβγαβγ

( ) ( ) ( ) 010103101023

=−−− vvv

0131010023

=−−−⇒ vvv

M1A1

M1

A1

A1

B1

M1A1

M1

A1

5

5 [10]

8 (i)

(ii)

Alternatively:

For final 3 marks in (i):

Award M1 for an attempt at formulae for all three coefficients. A1 for any two correct. A1 for completion

For final 4 marks in

(ii):

Award M1 for an attempt at formulae for all three coefficients. A1 for any one correct. A1 for a second one correct. A1 for completion.

Let equation be u3 + bu2 + cu + d = 0. –b = ∑α = 1⇒b = –1 c = 4∑αβ – (∑α)2 = 4 × (–3) – 12 = –13

( ) αβγααβα 843

−−=− ∑∑∑d

= 4 × 1 × (–3) – 13 – 8 × 10 = –93 So u3 – u2 – 13u + 93 = 0 Let equation be v3 + bv2 + cv + d = 0.

10

1

10

1−=⇒=

Σ=− bbαβγ

α

100

3

10

3

)( 22−=

−=

Σ=

αβγ

αβc

100

1

10

1

)(

122

−=⇒==− ddαβγ

So 0100

1

100

3

10

1 23=−−− vvv

or 100v3 – 10v2 – 3v – 1 = 0.

(5)

(5) [10]




Qu

No


Mark

Total

9 Finds normal vector to plane.

Uses known point to find constant term.

Angle between normals is equal to angle between planes.

Solve plane equation simultaneously. (Note: may find direction from vector product and use with one point.)

=

−

−−

4

1

6

221

121

kji

or

µλ

µλ

µλ

21

223

2

−−=

+−−=

++=

z

y

x

Plane equation is 6x + y + 4z = constant. Substitute (2, –3,1)⇒12 – 3 + 4 = 13. Or eliminate λ and µ. ⇒Π1 : 6x + y + 4z = 13

2253

4

323416

)k3j2i3).(k4ji6(cos

222222

=

++++

−−++=θ

=⇒θ 83.3º or 1.45 rad.

6x + y + 4z = 13 and 3x – 2y – 3z = 4 Obtains e.g. y + 2z = 1 and 3x + z = 6 Or two of (0,–11,6) , (11/6,0,1/2) , (2,1,0)

−

−

+

=

3

6

1

0

1

2

t

z

y

x

(OE)

Alternatively:

Direction of line from vector product. Finds a point on line. States equation of line.

M1A1

M1

A1

M1A1 A1

M1

A1A1

A1

(M1A1) (A1) (A1)

4

3

4

[11]




Qu

No


Mark

Total

10

(i)

(ii)

Reduces augmented matrix

Obtains set of values for a, giving unique solutions. Case of no solutions.

Case of infinite solutions.

Obtains particular solution.

−

−

+

−

−

−

−

−

−

−

−

−

−

−

−

8

3

7

62

6

2

0

5

2

0

0

1

~

29

11

7

2

10

2

6

9

2

3

2

1

a

a

a

a

Unique solution for all real a except a = – 3 or 5 Alternatively for first two marks:

0)3)(5(0

263

10)9(2

221

≠+−⇒≠

−

−−

−−

aa

a

a

a = –3 : 2a + 6 = 0 ⇒−=⇒ 80z no solutions (AG)

2

1 :5 −=⇒= za and x – 2y = – 8 (*)

∴infinite number of solutions (AG)

2

1−=z and x + y + z = 2⇒ x + y =

2

5

Solving simultaneously with (*) gives

x = –1 y = 2

7 z = –

2

1

M1A1

A1A1

M1A1√

M1A1√

B1

M1A1

4

2

2

3 [11]




Qu

No


Mark

Total

11 EITHER

Uses x2 + y2 = r2, x = r cos θ and y = r sin θ. One mark for each loop, or half of whole curve. Uses sector area formula. S.C. Omission of ½ factor, but correct integration gets B1. Differentiates. Puts 0=′y . Obtains coordinates.

(x2 + y2)2 = a2(x2 – y2)

−=⇒

2

2

2

2

22

r

y

r

xar

= a2 (cos2 θ – sin2 θ) = a2 cos 2θ (AG) Sketches C.

Area = ∫ ∫−

=4

4

4

0

22d 2cosd 2cos

2

1π

π

π

θθθθ aa

4

0

2

2

2sin

π

θ

= a

2

2a

=

)22()22)((2 222yyxayyxyx ′−=′++

xayxxy222 )(20 =+⇒=′

)0( 2

2 22≥=⇒=⇒ r

a

rar

2

12cos =⇒ θ

6

πθ ±=⇒ ,

6

5π±

i.e.

±6

,2

πa

and

±

6

5,

2

πa

M1

A1

B2,1,0

M1

A1A1

B1B1 M1

A1

M1

A1A1

2

2

3

7

[14]




Qu

No


Mark

Total

Alternatively for last 7

marks:

Obtains condition for tangent parallel to initial line. Differentiates equation of C. Forms equation. Solves for tanθ,

and θ. Writes coordinates of points.

0sind

dcos0

d

d0

d

d=+⇒=⇒= θ

θθ

θ

rr

y

x

y

θθ

θθ

2sind

d2sin2

d

d2

2

2

r

ar

a

r

r −=⇒−=

0sin2sincos

2

=−∴ θθθr

a

r

θθθθ sin2sincos2cos

22aa =∴

θθ

tan2tan

1=∴

13122

1 222

2

=⇒−=⇒=−

∴ tttt

t

t

3

1±=∴ t

6

5 ,

6

ππθ ±±=∴

±6

,2

πa

,

±

6

5,

2

πa

(Award final A1 if r found but result not written out.)

(M1A1)

(A1)

(M1)

(M1)

(A1)

(A1)

(7)

[14]




Qu

No


Mark

Total

11 OR

Differentiates once.

Differentiates again.

Substitutes.

Obtains result Finds CF. Finds PI.

Finds GS.

Evaluates coefficients from initial conditions.

Finds particular solution.

x

zxz

x

y

d

d

d

d+=

2

2

2

2

d

d

d

d2

d

d

x

zx

x

z

x

y+=

+−+−+

−++

222

226

926

d

d26

2

d

d

x

z

x

z

z

x

z

x

z

x

z

xxx

z

x2sin169=

xz

x

z

x

z

2sin1699d

d6

2d

2d=++⇒ (AG)

(Mark similarly if substitution is rearranged to x

yz = .)

30)3(96 22

−=⇒=+=++ mmmm CF: Ae–3x + Bxe–3x

PI: y = p sin 2x + q cos 2x y′ = 2p cos 2x – 2q sin 2x y″ = –4p sin 2x – 4q cos 2x

169125 =− qp 0512 =+ qp

⇒ p = 5 and q = – 12 GS: z = Ae–3x + Bxe–3x + 5 sin 2x – 12 cos 2x

21210 =⇒−=− AA z′ = –6e–3x + Be–3x – 3Bxe–3x + 10 cos 2x + 24 sin 2x

11065 =⇒++−= BB

xxxzxx

2cos122sin5ee233

−++=−−

xxxxxxy

xx

2cos122sin5ee2323

−++=∴−−

B1

B1

M1

A1

M1 A1

M1

M1 A1

A1

B1 M1 A1

A1

4

10 [14]

Documents

MARK SCHEME for the May/June 2012 question paper for the ...maxpapers.net/CIE/A Level/Mathematics – Further (9231)/9231_s12... · UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS